Chapter 11 - Thermochemistry
Chapter 11: Thermochemistry
– Heat and Chemical Changes
Part 1 – Notes: Enthalpy and Bond Energies
Objectives: Explain the relationship between
energy
and heat.
Identify, define, and explain:
thermochemistry,
energy, heat, endothermic, exothermic, chemical potential energy.
Determine the approximate
enthalpy of
a formation reaction with bond energies.
Text Reference: Section 11.1 – pages 293-299
&
part of Section 11.2 – pages 300-306
Energy:
Work (W):
Heat (q):
Temperature (T):
Every sample of matter has energy stored in it. From where does
this stored energy in matter come?
The total of all these forms of energy is called heat content, or the ENTHALPY, of the
substance. The symbol for enthalpy is H.
Enthalpy (H):
In all chemical reactions, there is a change in enthalpy (∆H).
The total enthalpy of the products is different from the total enthalpy
of the reactants. Remember,
change in enthalpy is the enthalpy of the final products minus the
enthalpy of the reactants.
Endothermic:
In order for a
process to absorb heat, what must be true of the initial
reactant versus the final products?
What is true
of the change in enthalpy of an endothermic process?
Exothermic:
In order for a
process to release heat, what must be
true of the initial reactant versus the final products?
What is true
of the change in enthalpy of an exothermic process?
Heat of reaction
(ΔH):
Molar heat of
formation (∆Hf):
Compounds with positive or low
negative values of heats of formation are generally . . .
Example: ΔHf of
H2S = -4.82
kcal/mol; ΔHf of HI =
+6.33 kcal/mol
Compounds with high negative
values of heats of formations are very . .
.
Example: ΔHf of
CO2 = -95.05
kcal/mol
Compounds with high positive
values of heats of formation are very . . .
Example: ΔHf of HgC2N2O2 (mercury
fulminate) = 64 kcal/mol
The standard enthalpy of formation (∆Hfo)
of any element in its standard state is ZERO.
Standard conditions for thermodynamic processes are generally 1 atm of
pressure and 25oC.
Standard
conditions are indicated by placing a o next to the
quantity.
∆Hf = heat
of formation but ∆Hfo
= standard heat of formation
Heat of combustion (ΔHc):
Enthalpy changes
for reactions can be obtained by simply subtracting the heats of
formation of the reactants from the heats of formation of the
products. Be sure to multiply the heats of formation by the
coefficient of the compound involved.
ΔHreaction = Σ ΔHproducts
- Σ ΔHreactants
Example 1:
Find the enthalpy (ΔHro)
for the reaction: CuO(s)
+ H2 (g)
→ Cu(s) +
H2O(g)
It is important to remember to specify the state of a substance when
its ∆Hf is
listed since different states of the same substance have different heat
contents. Remember, heat is absorbed or released when a substance
changes state: H2O(s) → H2O(l)
∆H = 6.01 kJ
How do compounds have energy of formation? If elements have no
energy of formation, how does a compound have an energy?
Bond dissociation energy:
In order to
break bond . . .
When bonds are
formed . . .
Let’s look at some bond dissociation energies:
H –
H
436.4 kJ
O =
O
498.7 kJ
H -
C
414 kJ
H –
O
460. kJ
C –
C
347 kJ
C =
C
620. kJ
Example 2: Ethene Hydrogenation:
What is the approximate ∆Hro
for the following reaction: C2H4
+ H2
→ C2H6?
From where does the
∆Hro
for this reaction come? Most of the energy of the reaction comes
from its bond energies.
Example 3:
Calculate the approximate ∆Hfo
for the formation of water: 2 H2 + O2
→ 2 H2O
Question:
How do you designate standard
conditions?
Question:
What are standard conditions?
Chapter 11: Thermochemistry – Heat
and
Chemical Changes
Part 1 – Assignment: Enthalpy and Bond Energies
Answer the following questions.
1. When chemical bonds are broken, energy is
_______________ but when they are formed energy is _______________.
2. For exothermic reactions, ΔH values
are
(+) / (–).
For endothermic reactions, ΔH values
are
(+) / (–).
3. Distinguish between these various forms of energy:
chemical potential energy, work, and heat.
4. A system is a person and the system is next to a
campfire. Is this system endothermic or exothermic? Explain.
5. A system is a person who is perspiring. Is
this system endothermic or exothermic? Explain.
6. Circle the letter(s) of the true statements below.
a. Energy is
detected only because
of its effects.
b. Heat is energy
that transfers from one object to another because they are at the sane
temperature.
c. Heat flows
from a cooler object
to a warmer object.
d. If two
objects remain in contact, heat will flow from the warmer to the cooler
until their temperatures are the same.
e. When a substance
dissolves in water, heat is always released.
f. The sign of
∆H is negative for an exothermic reaction.
g. If 129 kJ of heat is
required to decompose 2 mol NaHCO3, then 258 kJ is required
to decompose 1 mole.
h. In endothermic
reactions, the energy of the product(s) is higher than the energy of
the reactants.
7. The value for ΔHfo
for magnesium nitride is listed
as –461 kJ/mol. This means that the enthalpy of 1 mole of the Mg3N2
is 461 kJ GREATER /
LESS than the sum
of the enthalpies of Mg and N2
8. The value of ΔHfo
for ethane is listed as 52.3 kJ/mol. This means that the enthalpy
of 1 mole of C2H6
is 52.3 kJ GREATER /
LESS than the sum of the enthalpies of C and H2.
9. For the reaction: 2 HgO(s) → 2
Hg(s) + O2
(g) the ΔHr is +43.4
kcal. What is the ΔHf
for HgO?
10. The ΔHc
for one mole of ethyl alcohol is -227 kcal. How much heat is
evolved when 11.5 g of ethyl alcohol is combusted?
First you need to write a reaction…
11. The following are some ΔHf
values: CH4(g)
= -17.9 kcal/mol; CHCl3(l) =
-31.5 kcal/mol; HCl(g) = -22.1 kcal/mol
What is the heat of reaction for the equation:
CH4(g) + 3 Cl2(g)
→ CHCl3(l) + 3 HCl(g)
12. Calculate the heat of reaction for the
decomposition of sodium chlorate:
NaClO3(s) → NaCl(s) +
3/2 O2(g)
ΔHf for
NaClO3 = -85.7 kcal/mol; ΔHf
for NaCl = -98.2 kcal/mol
Chapter 11: Thermochemistry
– Heat and Chemical Changes
Part 2 – Notes: Thermochemistry and Hess’ Law (and a Little Entropy)
Objectives:
Identify, define, and explain: thermochemistry, system,
surroundings, universe, endothermic, exothermic, Hess’ Law, isobaric,
isothermal, and Hess’ Law.
Use Hess’ Law to calculate/determine the enthalpy
changes in chemical and physical processes.
Calculate enthalpy changes using standard heats of
formation.
Explain the relationship between
the system and the surroundings and the flow of heat between the two.
Text Reference:
Section 11.1 – pages 292-299 & Section 11.4 – pages 314-318
Introductory Question:
Natural systems tend to go from a state of
higher energy to a state of lower energy. What does this mean
with
respect to endothermic and exothermic reactions? Then, why do
endothermic
reactions take place at all?
Thermodynamics:
Recall, change in enthalpy is the change in the heat content of a
system at a constant pressure.
Isobaric:
Isothermal:
If you keep the temperature constant, then the change in energy of the
system is in the form of work on or by the system.
System:
Surroundings:
System +
Surroundings = Universe
Heat flowing
from the system to the surroundings is . . .
Heat flowing
from the surroundings to the system is . . .
What happens when two objects come in contact?
Is there another way to determine the energy of a reaction without
knowing the heats of formation for the components? Yes!!!
Hess’ Law: the enthalpy
change for a reaction is the sum of the enthalpy changes for a series
of reactions that add up to the overall reaction.
In other words: When
a reaction can be expressed as the algebraic sum of two or more other
reactions, then the enthalpy is the algebraic sum of the heats of
formation of these other reactions.
Example 1:
Find the enthalpy (ΔHr)
for the reaction: CuO(s)
+ H2 (g) →
Cu(s) +
H2O(g)
We can find: Cu(s) + 1/2 O2
(g) → CuO(s) ΔHf
= -37 kcal
H2(g)
+ 1/2 O2 (g) → H2O(l)
ΔHf
= -57.8 kcal
Example
2: Calculate the ΔH for the
reaction: 2 C(s) + H2(g)
→ C2H2(g)
Use the following reactions and their respective enthalpy
changes:
a)
C2H2(g)
+ 5/2 O2(g)
→ 2 CO2(g)
+ H2O(l)
ΔH
= -1299.6 kJ
b)
C(s) + O2(g)
→ CO2(g)
ΔH =
-393.5 kJ
c) H2(g)
+ 1/2 O2(g)
→ H2O(l)
ΔH =
-285.9 kJ
Example 3:
Liquid lead (IV) chloride is formed by the
reaction of solid lead (II) chloride with gaseous chlorine. Write
an
equation for this reaction. Then, calculate the ∆Hr.
Use the following reactions: Pb(s) + 2 Cl2(g)
→ PbCl4(l)
ΔH = -329.2 kJ
Pb(s) + Cl2(g)
→ PbCl2(s)
ΔH = -359.4 kJ
Just as a short introduction to the next topic, let’s have a quick
overview of a property called ENTROPY.
Entropy (S):
Something with perfect order would have no disorder and would have S =
0.
What would have ZERO ENTROPY?
Is there anything at these conditions?
What does this mean with regard to entropy values of
substance?
Think about the entropy of solids, liquids, and gases.
crystalline
solid
|
pattern:
|
degree of
order:
|
entropy
values:
|
liquids
|
pattern:
|
degree of
order:
|
entropy
values:
|
gases
|
pattern:
|
degree of
order:
|
entropy
values:
|
Is the entropy of a substance very significant and useful to us?
What is significant to us with regard to entropy?
How do you find the change in entropy?
ΔS = Σ ΔS(final
state) – Σ ΔS(initial state)
What if a reaction has a positive value for ∆S? What is an
example of such a process?
What is a reaction has a negative value for ∆S? What is an
example of such a process?
What do you think is the most likely thing to happen: increase in
entropy or decrease in entropy?
Chapter 11: Thermochemistry – Heat
and
Chemical Changes
Part 2 – Assignment: Thermochemistry and Hess’ Law
1. What can be considered the “system” and what are
considered the “surroundings” when studying a mixture of chemicals
undergoing a reaction? How is the flow of heat described?
2. A considerable amount of heat is required to is
required for the decomposition of aluminum oxide. (a) What is the
ΔHf for aluminum oxide?
(b) Is the reaction endothermic or exothermic?
Reaction: 2 Al2O3(s)
→ 4 Al(s) + 3 O2(g)
ΔH = 3352 kJ
3. Find the ΔHr
for the reaction
NO(g) + ½ O2(g)
→ NO2(g)
Use the following information:
½ N2(g)
+ ½ O2(g)
→ NO(g) ΔHf
= 90.4 kJ/mol
½ N2(g)
+ O2(g) → NO2(g)
ΔH2
= 33.6 kJ/mol
4. Find the ΔHr
for the reaction 2 P(s) + 5 Cl2(g)
→ 2 PCl5(s)
Use the following information: PCl5(s)
→ PCl3(g) +
Cl2(g)
ΔH = 87.9 kJ
2 P(s)
+ 3 Cl2(g) →
2 PCl3(g) ΔH
= -574 kJ
5. What is the ΔHf
for PCl5(s) from the above
problem?
6. The combustion for ethene is as
follows: C2H4(g)
+ 3 O2(g) →
2 CO2(g) + 2
H2O(l)
ΔH = -1390 kJ
Calculate the amount of heat liberated when 4.79 g
ethane reacts with excess oxygen.
7. Find the ΔHr
for the reaction C2H4(g)
+ H2(g) → C2H6(g)
Use the following information: 2 H2(g)
+ O2(g) → 2 H2O(l)
ΔH = -572 kJ
C2H4(g)
+ 3 O2(g) →
2 H2O(l) + 2
CO2(g)
ΔH = -1401 kJ
2 C2H6(g)
+ 7 O2(g) → 6 H2O(l)
+ 4 CO2(g)
ΔH = -3100.kJ
8. Diborane, (B2H6)
is a highly reactive boron hydride, which was once considered as a
rocket fuel in the U.S. space program. Calculate ΔH for the
synthesis of diborane from its elements, according to the
equation: 2 B(s) + 3 H2(g)
→ B2H6(g)
Use the following data:
(a) 2 B(s) +
3/2 O2(g)
→ B2O3(s)
ΔH = 01273 kJ
(b) B2H6(g)
+ 3 O2(g)
→ B2O3(s)
+ 3 H2O(g)
ΔH = -2035 kJ
(c) H2(g)
+ 1/2 O2(g)
→ H2O(l)
ΔH =
-286 kJ
(d) H2O(
l ) → H2O(g)
ΔH
= 44 kJ
Chapter 11: Thermochemistry
– Heat and Chemical Changes
Part 3 – Notes: Entropy, Free Energy, & Spontaneity
Objectives:
Identify, define, and explain: free energy, spontaneity, entropy,
nonspontaneous reaction, spontaneous reaction, law of
disorder, and Second Law of Thermodynamics.
Calculate the change in
entropy of a reaction or process and explain its significance with
regard to spontaneity.
Calculate the free energy
for a reaction or process and explain its significance to spontaneity.
Identify and define
conditions where a reaction will
be spontaneous, nonspontaneous, or in equilibrium.
Describe how
enthalpy change and entropy change may work together or against one
another in conjunction with temperature to determine the spontaneity or
nonspontaneity of a reaction.
Text Reference:
Section 19.3 – pages 549-556
Introductory Questions:
What is entropy?
What does a positive value of ∆S indicate?
What does a negative value of ∆S
indicate?
What is the more likely
situation: an increase in entropy or a decrease in entropy?
Second Law of Thermodynamics:
There are many versions of the second law, but all describe the
directionality of the universe
Version 1: It
is impossible to completely convert heat into work without some other
changes in the universe.
Version 2:
Heat will not of itself flow from a colder to a hotter body.
Version 3: The most general statement of the
second law of thermodynamics:
The entropy of the universe is increasing.
Sock Drawer Example:
Sock drawers do get organized and water can be decomposed into hydrogen
and oxygen and refrigerators transfer heat from a colder to a hotter
body. All of these are “unnatural” events – nonspontaneous in the vocabulary of
thermodynamics. They will not
occur by themselves; they require that work be done by someone or
something. An input of energy is necessary to reduce the
entropy and increase the order. And in every case, the work that
is done generates more entropy somewhere in the universe than it
reduces in one small part of the universe. Even when entropy
appears to decrease in a spontaneous
change that occurs “by itself,” for example, the freezing of water at
temperature below 0oC, there are balancing increases in entropy.
In this particular case, the heat given off by the freezing water adds
to the disorder of its surroundings. In short, when the entire
universe is considered, entropy always increases.
Spontaneous reaction:
Nonspontaneous reaction:
In terms of ENTHALPY:
spontaneous/favorable:
nonspontaneous/unfavorable:
In terms of ENTROPY:
spontaneous/favorable:
nonspontaneous/unfavorable:
Do endothermic reactions occur? YES. Do reactions happen
where there is a decrease in entropy? YES!
So how do these two properties come together to determine if a reaction
will be spontaneous or not?
Free Energy or Gibbs Free Energy:
the energy of a system that is available to do work
This relationship between enthalpy and entropy was discovered by
American scientist William Gibbs, and he showed the effect of entropy
depends upon the temperature. The Gibbs Equation
is:
ΔG = ΔH – T ΔS
ΔH = change in enthalpy; T =
temperature in kelvin; ΔS = change in entropy
The Gibbs Free Energy Equation shows that the TΔS of a system is
equivalent to the amount of heat that must be transferred to produce a
given change of
entropy, ΔS, at the temperature T.
Gibbs showed that a change is spontaneous ONLY if the change in free
energy, ΔG, is NEGATIVE.
It is not enough for ΔH to be NEGATIVE for a change to be
spontaneous. It is not enough for the ΔS to be POSITIVE for a change to be
spontaneous.
For a reaction to
be spontaneous, the combination of ΔH and –T ΔS must be NEGATIVE,
indicating a release of total free energy.
For a reaction to be spontaneous, the change in the free energy must be
. . .
For a reaction that is nonspontaneous, the change in free energy is . .
.
If the change in free energy is ZERO, . . .
Example 1:
For a given reaction, ΔH = -37 kcal, ΔS = -0.060 kcal/K, and T =
300oC. Calculate the ΔG at 300oC.
Since ΔG is
_______________, the change _______________ take place spontaneously at
300oC.
Example 2:
Use the above values for the reaction occurring at a temperature
of 400oC. Calculate ΔG.
Since ΔG is
_______________, the change _______________ take place spontaneously at
400oC.
KEY NOTE:
The whole notion of free energy tells us that there are two factors
that favor a reaction: a decrease in energy in the form of enthalpy and
an increase in disorder. If these two factors are working against
each other, then temperature will determine which is the more dominant
factor – the higher the temperature, the more entropic considerations
override enthalpic ones. Note: Also, the implication that a
reaction that is favorable at one temperature may not be favorable at
another.
Because the temperature is always positive, i.e., in Kelvin, the
effects of the signs of ΔH and ΔS and the effect of spontaneity can be
summarized in the following table.
Let’s examine four different situations.
Situation
|
Signs og
enthalpy and entropy
|
Comment
|
1
|
ΔH
= negative (favorable)
ΔS = positive (favorable)
|
|
2
|
ΔH
= positive (unfavorable)
ΔS = negative (unfavorable)
|
|
3
|
ΔH
= negative (favorable)
ΔS = negative (unfavorable)
|
|
4
|
ΔH
= positive (unfavorable)
ΔS = positive (favorable)
|
|
Chapter 11: Thermochemistry – Heat
and
Chemical Changes
Part 3 – Assignment: Entropy, Free Energy, & Spontaneity
Complete the assignment on a separate sheet of paper. You do not
have enough room to squeeze your work into the space here.
1. For a certain reaction, ΔH = -22 kcal and ΔG = -12
kcal at 25°C. (a) Calculate the ΔS. (b) For the same
reaction,
calculate ΔG when the temperature is 227°C.
2. The ΔH for a certain reaction is –30.0 kcal and
the ΔS is 0.080 kcal/K. (a) Calculate the ΔG at 25°C.
(b) Will this reaction occur spontaneously at 500. K? Explain how
you know. (c) Is this an endothermic or exothermic
reaction? Explain how you know.
3. Explain, by calculating ΔG, why KCl crystals will
dissolve in water at 25°C, although the ΔH is positive. (You
need to use some words in your explanation to indicate why the ΔG
is significant.) NOTE: ΔH = +2.0 kcal; ΔS = +0.023 kcal/K
4. For the phase change, H2O(s)
→ H2O(l),
th e value of ΔH is 1440 cal/mole. The ΔS value is 5.26 cal/mol
K. Calculate ΔG and state whether the change is spontaneous at
(a) +10.0°C and (b) –10.0°C. (c) Is the reaction
endothermic or exothermic? Explain how you know.
5. Determine whether the following reaction is
spontaneous: C(s) (graphite) + O2
(g) → CO2 (g)
ΔH = -393.5 kJ/mol
ΔS = +0.0030 kJ/K mol T =
298.15 K
6. Predict the sign of ΔS° for each of the
following reactions:
a. The thermal
decomposition of solid calcium carbonate: CaCO2
(s) → CaO(s) + CO2
(g)
b. The
oxidation of SO2 in
air: 2 SO2
(g) + O2 (g)
→ 2 SO3 (g)
7. Calculate the ΔS° at 25°C for the
reaction: 2 NiS(s) + 3 O2
(g) → 2 SO2
(g) + 2 NiO(s)
Use the following standard
entropy values:
SO2
(g) S° = 248 J/mol K
NiO(s) S° = 38 J/mol K
NiS(s) S° =
53 J/mol K O2
(g) S° = 205 J/mol K
8. Calculate ΔS° for the reduction of aluminum
oxide by hydrogen gas: Al2O3
(s) + 3H2 (g) → 2 Al(s) + 3 H2O(g)
Use the following standard entropy
values: Al2O3
(s) S° = 51 J/mol K
H2
(g) S° = 131 J/mol K
Al(s) S° = 28 J/mol
K
H2O(g) S° = 189 J/mol K
9. Consider the reaction: 2 SO2
(g) + O2 (g)
→ 2 SO2 (g) carried out
at 25°C and 1 atm. Use the following data:
SO2
(g) ΔH°
= -297 kJ/mol ΔS° = 248 J/mol K
SO3
(g) ΔH°
= -396 kJ/mol ΔS° = 257 J/mol K
O2
(g) ΔH° = 0
kJ/mol
ΔS° =
205 J/mol K
(a) Calculate ΔH°.
(b) Calculate ΔS°.
(c) Calculate ΔG°.
(d) Is the reaction endothermic or
exothermic? How do you know?
(e) Is the reaction spontaneous at
25°C? How do you know?
10. Given: a.
Na(s) + ½ Cl2
(g) → NaCl(s)
ΔH° = -93.23 kcal
b. H2
(g)
+ S(s) + 2 O2
(g) → H2SO4
(l) ΔH° = -193.91 kcal
c.
2 Na(s) + S(s) + 2 O2
(g) → Na2SO4
(s) ΔH° = -330.50 kcal
d.
½ H2 (g)
+ ½ Cl2 (g)
→ HCl(g)
ΔH° = -22.06 kcal
(a) Find the ΔHr for the following
chemical change: 2 NaCl(s) + H2SO4
(l) →
Na2SO4(s)
+ 2 HCl
(b)
Is the reaction endothermic or
exothermic? How do you know?
11. Calculate the enthalpy of combustion (ΔHc) for
the reaction in which ethane combines with oxygen to form carbon
dioixde and water.
Use the following information: 2
C(s)
+ 3 H2 (g) →
C2H6
(g)
ΔH = -20.2 kcal
C(s) + O2
(g) → CO2
(g)
ΔH
= -94.0 kcal
H2
(g) + 1/2 O2
(g) → H2O(g)
ΔH = -57.8 kcal
Chapter 11: Thermochemistry
– Heat and Chemical Changes
Part 4 – Notes: Calorimetry and Calculations
Objectives:
Identify, define, and explain: calorimetry, calorimeter,
enthalpy, thermochemical equation, heat of reaction, heat of
combustion, adiabatic, and isothermal.
Construct
equations that show heat changes for chemical and physical processes.
Calculate heat
changes and/or temperatures or temperature changes for chemical or
physical changes.
Text Reference:
Section 11.2 – pages 300-306
A calorimeter is a piece of equipment in which a reaction is performed adiabatically
(or isothermally)
meaning no heat flow is able to occur. The energy of the reaction
is kept inside the calorimeter and not allowed to escape into
the environment. Using the calorimeter, the heat involved in a
reaction
may be determined. This is how the calories (or Calories) of food
is
determined. Foods are combusted in calorimeters and their heat
content
is determined.
Often, a sample with a higher temperature is added to a material of
lower temperature. After the system equilibrates, the temperature
of the two
substances is the same. Many calculations are done based on this
premise.
When a “colder” substance is added to a “warmer” substance, certain
things
are true. Let’s point out some things. (pardon me if these
seem
obvious.)
1.
The “colder” substance will increase in temperature (heat energy
“flows” into it). The “warmer” substance will decrease in
temperature (heat energy “flows” out of it).
2.
The whole mixture will wind up at the SAME temperature that is higher than
the low temperature but lower than the high temperature.
3.
The energy that was lost by the warmer substance (“flowed” out of
the warmer temperature) is equal in energy but opposite in sign to the
energy that was absorbed by the cooler substance (“flowed” in to the
cooler
substance).
Let’s look at a typical problem that would be done and how to take
these things into account.
Example 1:
A piece of iron with a mass of 21.5 g at a
temperature of 100.oC is dropped into a calorimeter containing 132 g of
water
at an initial temperature of 20.0oC. The specific heat
of iron is Cp
= 0.448 J/goC. The
specific heat of water is 4.184 J/goC.
What
is the final temperature of the system/mixture?
Recall, the heat lost by the iron has the same
absolute value as the heat gained by the water.
Example 2:
Determine the final temperature when 18.0 g of ice at –10.0oC
mixes with 275.0 g of water at 60.0oC.
For water: Cfus = 334 J/g, Cp liquid = 4.184 J/goC,
and Cp ice = 2.060 J/goC.
Example 3:
32.2 g of water at 14.9oC
mixes with 32.2 g of water at 46.8oC.
What is the final temperature of the mixture? The specific heat
of water is 4.184 J/goC.
Chapter 11: Thermochemistry – Heat
and
Chemical Changes
Part 4 – Assignment: Calorimetry and Calculations
Solve the following problems. Show all work, set-ups, units, etc.
1. 75.0 g of iron at a temperature of 67.0oC
is placed in a calorimeter containing 135 g of water at 32.0oC.
The specific heat
of water is 4.184 J/goC.
The specific heat of iron is 0.448 J/goC.
What is the final temperature of the system?
2. 250. g of water is in a calorimeter at an initial
temperature of 28.0oC.
45.0 g of X is added to the water. X has an initial temperature
of 73.0oC. The specific
heat of X is 1.87 J/goC.
The
specific heat of water is 4.184 J/goC.
What is the final temperature of the system? (Assume no reaction
occurs.)
3. A calorimeter contains 325 g of water at an
initial temperature of 22.0oC.
Water has a specific heat of 4.184 J/goC.
A sample of M is at 60.0oC and
is added to the water in the calorimeter. M has a specific heat
of 1.67 J/goC. The final
temperature of the system is 38.0oC.
What is the mass of the M?
4. A calorimeter contains 275 g of water at a
temperature of 82.0oC. A
35.0 g lump of Z is added to the water. The sample of Z is at an
initial temperature of 20.0oC.
No reaction occurs but, after no more changes occur, the final
temperature of the system is recorded as 51.0oC.
The specific heat of water is 4.184 J/goC.
What is the specific heat of Z?